Looking for an elegant proof of $\det(A) = \det(A^t)$ without Schur decomposition Looking for an elegant proof of $\det(\textbf{A}) = \det(\textbf{A}^{t})$ without Schur decomposition.
Proof 1 with Schur decomposition
$$\textbf{A} = \textbf{P}^{t}\Delta\textbf{P} \implies\textbf{A}^{t} = (\textbf{P}^{t}\Delta\textbf{P})^{t} = \textbf{P}^{t}\Delta^{t}\textbf{P}$$
So, $\textbf{P}$  is unitary matrix, $\textbf{P}^{t}=\textbf{P}^{-1}$.
$$\det(\textbf{P})=\det(\textbf{P}^{t})= \det(\textbf{P}^{-1})$$
 $\Delta$ is upper triangular matrix.
$$\det(\Delta)=\det(\Delta^{t}) \implies   \det(\textbf{A})=\det(\textbf{A}^{t})$$ 
 A: Here's one proof, not sure if it's elegant or not.
Clearly $A$ is invertible if and only if $A^\mathrm{T}$ is, so we limit our proof to invertible matrices. Invertible matrices are products of elementary matrices and the determinants of elementary matrices are easily verified to be invariant under transpose. Hence the result follows.
A: $\lambda$ is an eigenvalue of $A$ with $v$ as its non-zero eigenvector then
$$
(A-\lambda I)v=0\implies v^T(A-\lambda I)^T=0\implies \det(A^T-\lambda I)=0
$$
Therefore the eigenvalues of $A$ and $A^T$ are the same and they have the same geometric multiplicity. For those class of matrices with the same algebraic and geometric multiplicities, we can say that  the eigenvalues of $A$ and $A^T$ are the same in multiplicity too. 
Again independently we can show that the determinant of $A$ is the product of its eigenvalues and hence:
$$
\det A=\prod_{i=1}^n\lambda_i=\det A^T.
$$
So we use the fact that $A$ and $A^T$ have the same eigenvalues to prove the result. Note that one should not use $\det A=\det A^T$ to prove that $A$ and $A^T$ have the same eigenvalues to avoid circularity.
A: I'm not sure this is quite what you're looking for, but: 
For any map $f:V \to V$ with $n = \dim V$ finite, the induced map $\bar{f}:\bigwedge V^n \to \bigwedge V^n$ is a map on a one-dimensional space, and thus must be given by $\overline{f}(x) = \alpha x$ for some constant $\alpha$ (independent of a choice of basis); this constant $\alpha$ is by definition $\det f$. With the notation above, we have $\overline{A^t} = (\overline{A})^t$. It's then easy to verify directly that $\det(A^t) = \det(A)$.
A: Every square matrix $A$ has an LDU-factorization. That is if $A$ is a square matrix there is a lower-triangular matrix $L$, a diagonal matrix $D$, and an upper-triangular matrix $U$ such that $A=LDU$ and both $L$ and $U$ have 1s on their diagonals.
From this factorization, it is clear that $$\det(A)=\det(LDU)=\det L\cdot\det D\cdot\det U=\det U^T\cdot\det D^T\cdot\det L^T=\det(U^TD^TL^T)=\det(A^T)$$
The existence of this factorization can be derived in a way similar to how one finds the inverse of a matrix. Given $A$ construct the matrix $[A|I]$ and perform row reduction to make $A$ into an upper diagonal matrix. The resulting matrix on the right is the inverse of the matrix you want in the factorization. To get the upper diagonal matrix do the same thing with the row reduced matrix $A'$ on top of the identity matrix and perform column reduction. The resulting matrix on the bottom will be the inverse of matrix you want in the factorization.
